Answer to Question #146676 in Physical Chemistry for Sanaa

Question

Calculate the pH (giving your answer to 2 d.p.) of the following weak acids.

(a) 0.100 mol dm–3 methanoic acid (HCOOH) with Ka = 1.77 × 10–4 mol dm–3

(b) 0.200 mol dm–3 ethanoic acid (CH3COOH) with Ka = 1.76 × 10–5 mol dm–3

(c) 0.025 mol dm–3 pentanoic acid (C4H9COOH) with a pKa = 4.86

(d) 44 g of butanoic acid (CH3CH2CH2COOH), Ka= 1.51 × 10–5 mol dm–3, which was

dissolved in 2 dm3 of water.

Answer 1

A. Molar concentration of HCOOH= 0.100M

  Ka = 1.77 x 10-4

  HCOOH ::::::: H+ + HCOO_

 The Equilibrium concentration are

HCOOH = 0.100 - x

H+ = x

HCOO_ = x

Ka = [H+][HCOO_]

     [ HCOOH]

1.77 x 10_4 = (x)(x)

           0.100-x

1.77 x 10-4 =.  x2

                0.100-x

Since the acid is weak, 0.100-x is approximately 0.100

 X² = 1.77 x 10-4

0.100

x² = 1.77 x 10-4 x 0.100

x² = 1.77 x 10-5

x = 4.21 x 10-3

H+ = 4.21 x 10-3 

pH = - log [H+]

pH = - log [ 4.21 x 10-3]

pH = 2.375

[pH = 2.38]

B. CH3COOH = 0.200M

Ka = 1.76 x 10-5

The Equilibrium concentration are

CH3COOH = 0.200 - x

H+ = x

CH3COO- = x

Ka = [H+][CH3COO- ]

     [CH3COOH]

  x² = 1.76 x 10-5

0.200-x

Since the acid is weak, 0.200-x is approximately 0.200

  x² = 1.76 x 10-5

0.200

X² = 3.52 x 10-6

x = 1.876 x 10-3

[H+] = 1.876 x 10-3

pH = - log [ 1.876 x 10-3]

pH = 2.7267

[pH= 2.73]

C. C4H9COOH = 0.025M

pKa = 4.86

pKa = -log Ka

Ka = 10-4.86

Ka = 1.38 x 10-5

The Equilibrium concentrations are

 C4H9COOH = 0.025-x

H+ = x

 C4H9COO- = x

Ka = [H+][ C4H9COO-] 

     [ C4H9COOH]

  x² = 1.38 x 10-5

0.025-x

Since the acid is weak, 0.025-x is approximately 0.025

x² = 1.38 x 10-5

0.025

x² = 3.45 x 10-7

x = 5.87 x 10-4

H+ = 5.87 x 10-4

pH = - log [5.87 x 10-4]

[pH = 3.23]

D. Mass concentration of butanoic acid = 44g/2dm³ = 22gdm-3

  Molar mass of butanoic acid = 88gmol-

Molar concentration = 22/88 = 0.25moldm-3

Ka = 1.51 x 10-5

The Equilibrium concentrations are

C3H7COOH = 0.25-x

H+ = x

C3H7COO- = x

Ka = [H+][C3H7COO- ]

     [C3H7COOH]

 x² = 1.51 x 10-5

0.25-x

Since the acid is weak, 0.25-x is approximately 0.25

  x² = 1.51 x 10-5

0.25

x² = 3.775 x 10-6

x = 1.94 x 10-3

[H+] = 1.94 x 10-3

pH = -log [H+]

pH = - log [ 1.94 x 10-3]

[pH = 2.71]